An object weighing 98N is dropped from a height of 10m. It is found to be moving with a velocity 12m/s just before it hits the ground. How large was the frictional force acting upon it?
Will it be done by using k.E=P.E+W in which -26N ANSWER is coming. If yes,than can it be solved via equations of motion (than f=ma)..

Both methods can be used.

But answers are different?? I think 26N is the right answer which I am getting from work energy equation. But can't from equ of motion?

Show your work of how you reached 26 N please.

0.5mv^2-mgh=F.S
0.5*10*12^2 - (10*9.8*10)=f.s
-260N=F(10)
F=-26N..

Okay, how did you do for the equation of motion?

I am getting different answers.but not 26N.. Can u solve it please..

No no, I want you to post what you tried. There's nothing wrong with making a mistake. You learn from mistakes after all.

Here vi=0 or vf=0..

I won't answer anything more until you post what you tried, from the start to the final answer you got that you're saying is wrong.

Ive delivered my try before as well..
If vi=0 than,
f=ma
=10*(vf-vi)/t
=10*(12/1.428)
=84N.. WHERE t from 2nd equ of motion

Okay, I'm not sure why you tried to use something you have to calculate here.

I would use v^2 = u^2 + 2as

v = 12
u = 0
a = ?
s = 10

(12)^2 = 0^2 + 2a(10)

a = 7.2 m/s^2

This is the net acceleration of the object.

a_{net} = a_g - a_f

7.2 = 9.8 - a_f

a_f = 2.6 m/s^2

This is the acceleration due to friction.

F = ma

F = 10*2.6 = 26 N

Thanks mate..